Hard A-Level Simultaneous equations Questions

Equate line and curve

Intersection points satisfy both equations, so we set the $y$-expressions equal.

Form a quadratic in $x$

Rearranging gives a quadratic in $x$ whose real roots are the intersection points; its coefficients contain $k$.

Identify the coefficients

We list $a$, $b$, $c$ ready to build the discriminant.

Write the discriminant

For two distinct intersections we need the discriminant to be strictly positive.

Simplify the discriminant

Expanding gives a neat expression in $k$. We now show it is always positive.

Recognise a non-negative square

Writing it as a square plus a positive constant makes the sign obvious, because any real square is at least zero.

Deduce the minimum value

The smallest the expression can be is the positive constant, reached when the square is zero. So it is always positive.

Conclude two roots for all $k$

A strictly positive discriminant guarantees two distinct real roots, hence two intersection points, whatever the value of $k$.

Locate the vertex of the discriminant parabola

Viewing the discriminant as a parabola in $k$, its lowest point is where its rate of change is zero. This is where it is smallest.

Evaluate the minimum discriminant

Even at its smallest the discriminant is positive, so it is positive everywhere — the strongest form of the argument.

Interpret geometrically

Because the discriminant never reaches zero, the line is never a tangent and never misses — it always crosses the curve twice.

State the key idea

The whole proof rests on the fact that a real number squared cannot be negative, a technique used throughout A-Level inequalities and proof.

Confirm with a value

Checking a single value is consistent with the general proof and guards against slips.

Confirm with another value

A second check reinforces that the discriminant stays positive.

Write the conclusion carefully

A clear final sentence, quoting the discriminant being strictly positive, is what earns the proof marks.

Equate line and curve

Intersection points satisfy both equations, so we set the $y$-expressions equal.

Form a quadratic in $x$

Rearranging gives a quadratic in $x$ whose real roots are the intersection points; its coefficients contain $k$.

Identify the coefficients

We list $a$, $b$, $c$ ready to build the discriminant.

Write the discriminant

For two distinct intersections we need the discriminant to be strictly positive.

Simplify the discriminant

Expanding gives a neat expression in $k$. We now show it is always positive.

Recognise a non-negative square

Writing it as a square plus a positive constant makes the sign obvious, because any real square is at least zero.

Deduce the minimum value

The smallest the expression can be is the positive constant, reached when the square is zero. So it is always positive.

Conclude two roots for all $k$

A strictly positive discriminant guarantees two distinct real roots, hence two intersection points, whatever the value of $k$.

Locate the vertex of the discriminant parabola

Viewing the discriminant as a parabola in $k$, its lowest point is where its rate of change is zero. This is where it is smallest.

Evaluate the minimum discriminant

Even at its smallest the discriminant is positive, so it is positive everywhere — the strongest form of the argument.

Interpret geometrically

Because the discriminant never reaches zero, the line is never a tangent and never misses — it always crosses the curve twice.

State the key idea

The whole proof rests on the fact that a real number squared cannot be negative, a technique used throughout A-Level inequalities and proof.

Confirm with a value

Checking a single value is consistent with the general proof and guards against slips.

Confirm with another value

A second check reinforces that the discriminant stays positive.

Write the conclusion carefully

A clear final sentence, quoting the discriminant being strictly positive, is what earns the proof marks.

Equate line and curve

Intersection points satisfy both equations, so we set the $y$-expressions equal.

Form a quadratic in $x$

Rearranging gives a quadratic in $x$ whose real roots are the intersection points; its coefficients contain $k$.

Identify the coefficients

We list $a$, $b$, $c$ ready to build the discriminant.

Write the discriminant

For two distinct intersections we need the discriminant to be strictly positive.

Simplify the discriminant

Expanding gives a neat expression in $k$. We now show it is always positive.

Recognise a non-negative square

Writing it as a square plus a positive constant makes the sign obvious, because any real square is at least zero.

Deduce the minimum value

The smallest the expression can be is the positive constant, reached when the square is zero. So it is always positive.

Conclude two roots for all $k$

A strictly positive discriminant guarantees two distinct real roots, hence two intersection points, whatever the value of $k$.

Locate the vertex of the discriminant parabola

Viewing the discriminant as a parabola in $k$, its lowest point is where its rate of change is zero. This is where it is smallest.

Evaluate the minimum discriminant

Even at its smallest the discriminant is positive, so it is positive everywhere — the strongest form of the argument.

Interpret geometrically

Because the discriminant never reaches zero, the line is never a tangent and never misses — it always crosses the curve twice.

State the key idea

The whole proof rests on the fact that a real number squared cannot be negative, a technique used throughout A-Level inequalities and proof.

Confirm with a value

Checking a single value is consistent with the general proof and guards against slips.

Confirm with another value

A second check reinforces that the discriminant stays positive.

Write the conclusion carefully

A clear final sentence, quoting the discriminant being strictly positive, is what earns the proof marks.

Equate line and curve

As always, intersections occur where the $y$-values agree, so we set the expressions equal.

Rearrange to a quadratic in $x$

Collecting terms to one side gives a quadratic in $x$ whose coefficients involve $k$.

Identify the coefficients

We list $a$, $b$, $c$ to build the discriminant, which controls how many intersections there are.

Recall what the discriminant tells us

The discriminant decides the number of real solutions, which is the number of times the line meets the curve. This is the key idea linking algebra to the picture.

Form the discriminant

We substitute the coefficients in brackets first. Because $b$ contains $k$, the discriminant will be an expression in $k$.

Expand and simplify

Expanding gives the discriminant as a neat quadratic in $k$. This is the expression we now control with an inequality.

Apply the required condition

For the line to meets the curve as required, we impose the matching discriminant condition.

Find the critical values of $k$

First solve the related equation. These boundary values are where the line is exactly a tangent (one point of contact).

Sketch the quadratic in $k$

Thinking of the discriminant as its own parabola in $k$ shows where it is positive or negative without guessing.

Read off the solution set

The parabola is above the axis outside its two roots, so we take the two outer regions. Sketching the parabola makes choosing the correct region reliable.

Boundary tangent when $k=6$

At each critical value the line is a tangent; the single point of contact is $x=-b/2a$. This confirms the boundaries are exactly the tangent cases we exclude for 'two distinct points'.

Boundary tangent when $k=-6$

At each critical value the line is a tangent; the single point of contact is $x=-b/2a$. This confirms the boundaries are exactly the tangent cases we exclude for 'two distinct points'.

Test a value in the region

Substituting one value from the solution set back into the discriminant confirms we picked the correct region — a reliable check for quadratic inequalities.

Interpret graphically

Every real solution of the quadratic in $x$ is a point where the line meets the curve, so the discriminant condition matches the geometry exactly.

State the answer in set notation

Writing the final region clearly, using inequalities or set notation, is what the examiner is looking for.

Equate line and curve

As always, intersections occur where the $y$-values agree, so we set the expressions equal.

Rearrange to a quadratic in $x$

Collecting terms to one side gives a quadratic in $x$ whose coefficients involve $k$.

Identify the coefficients

We list $a$, $b$, $c$ to build the discriminant, which controls how many intersections there are.

Recall what the discriminant tells us

The discriminant decides the number of real solutions, which is the number of times the line meets the curve. This is the key idea linking algebra to the picture.

Form the discriminant

We substitute the coefficients in brackets first. Because $b$ contains $k$, the discriminant will be an expression in $k$.

Expand and simplify

Expanding gives the discriminant as a neat quadratic in $k$. This is the expression we now control with an inequality.

Apply the required condition

For the line to meets the curve as required, we impose the matching discriminant condition.

Find the critical values of $k$

First solve the related equation. These boundary values are where the line is exactly a tangent (one point of contact).

Sketch the quadratic in $k$

Thinking of the discriminant as its own parabola in $k$ shows where it is positive or negative without guessing.

Read off the solution set

The parabola is above the axis outside its two roots, so we take the two outer regions. Sketching the parabola makes choosing the correct region reliable.

Boundary tangent when $k=6$

At each critical value the line is a tangent; the single point of contact is $x=-b/2a$. This confirms the boundaries are exactly the tangent cases we exclude for 'two distinct points'.

Boundary tangent when $k=2$

At each critical value the line is a tangent; the single point of contact is $x=-b/2a$. This confirms the boundaries are exactly the tangent cases we exclude for 'two distinct points'.

Test a value in the region

Substituting one value from the solution set back into the discriminant confirms we picked the correct region — a reliable check for quadratic inequalities.

Interpret graphically

Every real solution of the quadratic in $x$ is a point where the line meets the curve, so the discriminant condition matches the geometry exactly.

State the answer in set notation

Writing the final region clearly, using inequalities or set notation, is what the examiner is looking for.